I understand how to take the derivative of an equation with respect to a variable, say S, but what do I do when there are two variables involved? Here's the problem d/dx [3cos(x^2) - 4e^t]

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- Mar 15th 2012, 11:15 AMNotAMathmatician314Tricky derivative problem!
I understand how to take the derivative of an equation with respect to a variable, say S, but what do I do when there are two variables involved? Here's the problem d/dx [3cos(x^2) - 4e^t]

- Mar 15th 2012, 11:23 AMKanwar245Re: Tricky derivative problem!
Treat t as constant, so

Quote:

d/dx [3cos(x^2) - 4e^t]

= 3 d/dx cos(x^2) - 4 d/dx e^t

= -6x sin(x^2) - 4e^t x

- Mar 15th 2012, 11:42 AMtom@ballooncalculusRe: Tricky derivative problem!
Correct. But then...

Quote:

, so

d/dx [3cos(x^2) - 4e^t]

= 3 d/dx cos(x^2) - 4 d/dx e^t

= 3 d/dx cos(x^2) - 4e^t d/dx 1

... but then the whole second term is a constant term and drops out...

= -6x sin(x^2)

... period.

As long as t isn't a function of x, or anything... the context of the problem here would help. - Mar 15th 2012, 11:54 AMNotAMathmatician314Re: Tricky derivative problem!
What would it look like if t was a function of x?

- Mar 15th 2012, 11:59 AMKanwar245Re: Tricky derivative problem!
my bad, I integrated the 2nd part!

- Mar 15th 2012, 01:27 PMskeeterRe: Tricky derivative problem!